42 2.2 Limit Theorems Therefore, limn →∞(an + bn)= a+ b. This proves (i). (ii) If k = 0, then ka = 0 and kan = 0 for all n. The conclusion follows immediately. Suppose next that k ̸ = 0. Given ε > 0, let N ∈ N be such that |an − a| < ε/|k| for n ≥ N. Then for n ≥ N, ε |kan − ka| = |k(an − a)| = |k||an − a| < |k| = ε. |k| It follows that limn →∞(kan)= ka as desired. This proves (ii). (iii) Since {an} is convergent, it follows from Theorem 2.1.4 that it is bounded. Thus, there exists M > 0 such that |an|≤ M for all n ∈ N. For every n ∈ N, we have the following estimate: |anbn − ab| = |anbn − anb + anb − ab|≤|an||bn − b| + |b||an − a|. (2.1) Let ε > 0. Since {an} converges to a, we may choose N1 ∈ N such that ε |an − a| < for all n ≥ N1. 2(|b| + 1) Similarly, since {bn} converges to b, we may choose N2 ∈ N such that ε |bn − b| < for all n ≥ N2. 2M Let N = max{N1,N2}. Then, for n ≥ N, it follows from (2.1) that ε ε ε |b| ε ε ε |anbn − ab| < M + |b| = + < + = ε for all n ≥ N. 2M 2(|b| + 1) 2 |b| + 12 22 Therefore, limn →∞ anbn = ab. This proves (iii). (iv) We frst show that 1 1 lim = . n→∞ bn b Since {bn} converges to b, there is N1 ∈ N such that |b| |bn − b| < for n ≥ N1. 2 Using Corollary (1.4.4) it follows that, for such n, |b| |b| |b| − < |bn|−|b| < and, hence, < |bn|. For each n ≥ N1, we have the following estimate 2 2 2 1 bn − 1 b = | bn − b| |bn||b| ≤ |bn − b| |b| | b| 2 = 2|bn − b| . b2 (2.2)
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